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# Linear Algebra Done Right

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of C as a vector space over C, then the list .1C i; 1 � i/ is linearly dependent. 6 Suppose v1; v2; v3; v4 is linearly independent in V. Prove that the list v1 � v2; v2 � v3; v3 � v4; v4 is also linearly independent. 7 Prove or give a counterexample: If v1; v2; : : : ; vm is a linearly indepen- dent list of vectors in V, then 5v1 � 4v2; v2; v3; : : : ; vm is linearly independent. 8 Prove or give a counterexample: If v1; v2; : : : ; vm is a linearly indepen- dent list of vectors in V and � 2 F with � ¤ 0, then �v1; �v2; : : : ; �vm is linearly independent. 9 Prove or give a counterexample: If v1; : : : ; vm and w1; : : : ;wm are lin- early independent lists of vectors in V, then v1 C w1; : : : ; vm C wm is linearly independent. 10 Suppose v1; : : : ; vm is linearly independent in V and w 2 V. Prove that if v1 C w; : : : ; vm C w is linearly dependent, then w 2 span.v1; : : : ; vm/. 38 CHAPTER 2 Finite-Dimensional Vector Spaces 11 Suppose v1; : : : ; vm is linearly independent in V and w 2 V. Show that v1; : : : ; vm;w is linearly independent if and only if w … span.v1; : : : ; vm/: 12 Explain why there does not exist a list of six polynomials that is linearly independent in P4.F/. 13 Explain why no list of four polynomials spans P4.F/. 14 Prove that V is inﬁnite-dimensional if and only if there is a sequence v1; v2; : : : of vectors in V such that v1; : : : ; vm is linearly independent for every positive integer m. 15 Prove that F1 is inﬁnite-dimensional. 16 Prove that the real vector space of all continuous real-valued functions on the interval Œ0; 1� is inﬁnite-dimensional. 17 Suppose p0; p1; : : : ; pm are polynomials in Pm.F/ such that pj .2/ D 0 for each j . Prove that p0; p1; : : : ; pm is not linearly independent in Pm.F/. SECTION 2.B Bases 39 2.B Bases In the last section, we discussed linearly independent lists and spanning lists. Now we bring these concepts together. 2.27 Deﬁnition basis A basis of V is a list of vectors in V that is linearly independent and spans V. 2.28 Example bases (a) The list .1; 0; : : : ; 0/; .0; 1; 0; : : : ; 0/; : : : ; .0; : : : ; 0; 1/ is a basis of Fn, called the standard basis of Fn. (b) The list .1; 2/; .3; 5/ is a basis of F2. (c) The list .1; 2;�4/; .7;�5; 6/ is linearly independent in F3 but is not a basis of F3 because it does not span F3. (d) The list .1; 2/; .3; 5/; .4; 13/ spans F2 but is not a basis of F2 because it is not linearly independent. (e) The list .1; 1; 0/; .0; 0; 1/ is a basis of f.x; x; y/ 2 F3 W x; y 2 Fg. (f) The list .1;�1; 0/; .1; 0;�1/ is a basis of f.x; y; z/ 2 F3 W x C y C z D 0g: (g) The list 1; z; : : : ; zm is a basis of Pm.F/. In addition to the standard basis, Fn has many other bases. For example, .7; 5/; .�4; 9/ and .1; 2/; .3; 5/ are both bases of F2. The next result helps explain why bases are useful. Recall that “uniquely” means “in only one way”. 2.29 Criterion for basis A list v1; : : : ; vn of vectors in V is a basis of V if and only if every v 2 V can be written uniquely in the form 2.30 v D a1v1 C � � � C anvn; where a1; : : : ; an 2 F. 40 CHAPTER 2 Finite-Dimensional Vector Spaces Proof First suppose that v1; : : : ; vn is a basis of V. Let v 2 V. Because v1; : : : ; vn spans V, there exist a1; : : : ; an 2 F such that 2.30 holds. To This proof is essentially a repeti- tion of the ideas that led us to the deﬁnition of linear independence. show that the representation in 2.30 is unique, suppose c1; : : : ; cn are scalars such that we also have v D c1v1 C � � � C cnvn: Subtracting the last equation from 2.30, we get 0 D .a1 � c1/v1 C � � � C .an � cn/vn: This implies that each aj � cj equals 0 (because v1; : : : ; vn is linearly inde- pendent). Hence a1 D c1; : : : ; an D cn. We have the desired uniqueness, completing the proof in one direction. For the other direction, suppose every v 2 V can be written uniquely in the form given by 2.30. Clearly this implies that v1; : : : ; vn spans V. To show that v1; : : : ; vn is linearly independent, suppose a1; : : : ; an 2 F are such that 0 D a1v1 C � � � C anvn: The uniqueness of the representation 2.30 (taking v D 0) now implies that a1 D � � � D an D 0. Thus v1; : : : ; vn is linearly independent and hence is a basis of V. A spanning list in a vector space may not be a basis because it is not linearly independent. Our next result says that given any spanning list, some (possibly none) of the vectors in it can be discarded so that the remaining list is linearly independent and still spans the vector space. As an example in the vector space F2, if the procedure in the proof below is applied to the list .1; 2/; .3; 6/; .4; 7/; .5; 9/, then the second and fourth vectors will be removed. This leaves .1; 2/; .4; 7/, which is a basis of F2. 2.31 Spanning list contains a basis Every spanning list in a vector space can be reduced to a basis of the vector space. Proof Suppose v1; : : : ; vn spans V. We want to remove some of the vectors from v1; : : : ; vn so that the remaining vectors form a basis of V. We do this through the multi-step process described below. Start with B equal to the list v1; : : : ; vn. SECTION 2.B Bases 41 Step 1 If v1 D 0, delete v1 from B . If v1 ¤ 0, leave B unchanged. Step j If vj is in span.v1; : : : ; vj�1/, delete vj from B . If vj is not in span.v1; : : : ; vj�1/, leave B unchanged. Stop the process after step n, getting a list B . This list B spans V because our original list spanned V and we have discarded only vectors that were already in the span of the previous vectors. The process ensures that no vector in B is in the span of the previous ones. Thus B is linearly independent, by the Linear Dependence Lemma (2.21). Hence B is a basis of V. Our next result, an easy corollary of the previous result, tells us that every ﬁnite-dimensional vector space has a basis. 2.32 Basis of ﬁnite-dimensional vector space Every ﬁnite-dimensional vector space has a basis. Proof By deﬁnition, a ﬁnite-dimensional vector space has a spanning list. The previous result tells us that each spanning list can be reduced to a basis. Our next result is in some sense a dual of 2.31, which said that every spanning list can be reduced to a basis. Now we show that given any linearly independent list, we can adjoin some additional vectors (this includes the possibility of adjoining no additional vectors) so that the extended list is still linearly independent but also spans the space. 2.33 Linearly independent list extends to a basis Every linearly independent list of vectors in a ﬁnite-dimensional vector space can be extended to a basis of the vector space. Proof Suppose u1; : : : ; um is linearly independent in a ﬁnite-dimensional vector space V. Let w1; : : : ;wn be a basis of V. Thus the list u1; : : : ; um;w1; : : : ;wn spans V. Applying the procedure of the proof of 2.31 to reduce this list to a basis of V produces a basis consisting of the vectors u1; : : : ; um (none of the u’s get deleted in this procedure because u1; : : : ; um is linearly independent) and some of the w’s. 42 CHAPTER 2 Finite-Dimensional Vector Spaces As an example in F3, suppose we start with the linearly independent list .2; 3; 4/; .9; 6; 8/. If we take w1;w2;w3 in the proof above to be the standard basis of F3, then the procedure in the proof above produces the list .2; 3; 4/; .9; 6; 8/; .0; 1; 0/, which is a basis of F3. Using the same basic ideas but considerably more advanced tools, the next result can be proved with- out the hypothesis that V is ﬁnite- dimensional. As an application of the result above, we now show that every subspace of a ﬁnite-dimensional vector space can be paired with another subspace to form a direct sum of the whole space. 2.34 Every subspace of V is part of a direct sum equal to V Suppose